University of California Los Angeles Calculus Analysis Questions

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Mathematics

University Of California Los Angeles

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Problem 1. Define fn: R+R by fn (2) = 1+ nr Show that fn converges uniformly to a differentiable function f and that f'(2) = limn+ f (2) whenever *#0, but f'0) # limn+tho). (Hint: for the uniform convergence, consider separately values of x such that 2 > 8 and such that |2 0.) Problem 2. Show that the series +n n2 n=1 converges uniformly on any bounded interval in R, but does not converge absolutely for any 2 ER. (Hint: For the uniform convergence, split the series as a sum of two series and prove uniform convergence for each of them.) Problem 3. Let (X,d) be a metric space, and E CX a subset. Show that the following conditions are all equivalent. (1) E= X (2) For every 2 EX, there exists a sequence (In)=1 in E such that limnyen = r. (3) For every nonempty open set ASX, AnE. Recall from lecture: if any of these equivalent conditions is satisfied, we say that E is dense in X. Problem 4. Suppose f : [a, b] + R is a continuous function. (a) Suppose f(x) > 0 for every x € [a, b] and ſo f(x) dx = 0. Show that f(x) = 0 for all & € [a, b]. (b) Suppose să f(x)2" dx = 0 for every n > 0. Show that f(1) = 0 for all r € [a, b]. (Hint for b: First, show that să f(x)P(x) dx = 0 for every polynomial P. Then use the Weierstrass approximation theorem for f to get that so f(x)2 dx = 0, and conclude using part (a). Note that Weierstrass approximation gives you uniform convergence. Do you need that in your proof?) Problem 5. (Not to be handed in.) Consider the set of trigonometric polynomials on [0, ļ] given by Ŝ | € 0, 3), N E N, , EC A 1- { ovations | 310 8.6} ? n=-N Show that A is dense in C([0, 1], C) by checking that all conditions of the Stone-Weierstrass theorem are satisfied. 1
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Explanation & Answer

View attached explanation and answer. Let me know if you have any questions.The solutions are ready, please look on them. I did what I could for q5, too.

First, determine the function 𝑓: clearly 𝑓(0) = 0 because 𝑓𝑛 (0) = 0 ∀𝑛 and 𝑓(𝑥) = 0 ∀𝑥 ≠ 0 because
for every 𝑥 ≠ 0 the numerator is fixed and the denominator tends to infinity.

|𝑥|

Now estimate |𝑓(𝑥) − 𝑓𝑛 (𝑥)| = |𝑓𝑛 (𝑥)| = 1+𝑛𝑥 2 separately for |𝑥| ≤

For 𝑥 ≥

1
√𝑛

1
√𝑛

and for 𝑥 ≥

1
√𝑛

.

we may estimate
|𝑥|
1
1
1
1
=


=
.
2
1
1
1 + 𝑛𝑥
√𝑛
+ 𝑛|𝑥| 𝑛|𝑥| 𝑛 ∙
|𝑥|
√𝑛

For |𝑥| ≤

1
√𝑛

use another estimate:
1
|𝑥|
1
√𝑛
=
=
.
1 + 𝑛𝑥 2
1
√𝑛

This way, for any 𝑥 and 𝑛 it is true that |𝑓(𝑥) − 𝑓𝑛 (𝑥)| ≤

1
√𝑛

. This estimate tends to zero independently

of 𝑥, so the convergence is uniform.

[where

1
√𝑛

comes from: it is the point where 𝑓𝑛 (𝑥) has a maximum]

It is clear that the series does not converge absolutely because
𝑥2 + 𝑛
𝑥2 + 𝑛
𝑛
1
𝑛
|(−1)
|
=

=
,
𝑛2
𝑛2
𝑛2 𝑛
which is a divergent series.

Uniform convergence:


𝑥2 + 𝑛
𝑥2
𝑛
𝑛
(−1)
(−1)𝑛 2
=

+

2
2
𝑛
𝑛
𝑛
𝑛=1
𝑛=1
𝑛=1
(we can write this way because both series on the right are convergent, too). Indeed, both series are
alternating series (signs are changed while absolute value decreases).




(−1)𝑛

1

𝑛
The second summand ∑∞
𝑛=1(−1) 𝑛 does not depend on 𝑥, so it converges uniformly (even on ℝ).
1

𝑛
2
The first summand is 𝑥 2 ∑∞
𝑛=1(−1) 𝑛2 . On any bounded interval 𝑥 is bounded by some 𝑀 while
𝑛
∑∞
𝑛=1(−1)

1

does not depend on 𝑥 and converges to some 𝑎. This way,
𝑁
1
𝜀
∀𝜀 > 0 ∃𝑁: ∀𝑛 ≥ 𝑁 |∑ (−1)𝑛 2 − 𝑎| < ,
𝑛
𝑀
𝑛=1
so for all 𝑥 in the interval
𝑁
𝑁
1
1
|𝑥 2 ∑ (−1)𝑛 2 − 𝑥 2 𝑎| ≤ 𝑀 |∑ (−1)𝑛 2 − 𝑎| < 𝜀,
𝑛
𝑛
𝑛=1
𝑛=1
which is what we need.
𝑛2

1

(𝟏) ⇒ (𝟐): by the definition of closure, every 𝑥 ∈ 𝑋 is an adherent point of 𝐸, so for every ball 𝐵 (𝑥, )
𝑛
there is a point 𝑥𝑛 ∈ 𝐸. Because

1

𝑛

→ 0, 𝑑(𝑥, 𝑥𝑛 ) → 0, too, so 𝑥𝑛 → 𝑥.

(𝟐) ⇒ (𝟑): suppose that 𝐴 ∩ 𝐸 = ∅, then some 𝑥 ∈ 𝐴. Because 𝐴 is open, some 𝐵(𝑥, 𝑟) ⊂ 𝐴 and
therefore has empty intersection with 𝐸. But this is a contradiction with (2) that some 𝑥𝑛 → 𝑥 from 𝐸.

1

(𝟑) ⇒ (𝟏): for any 𝑥 ∈ 𝑋, consider 𝐴𝑛 = �...


Anonymous
Just the thing I needed, saved me a lot of time.

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